6. Theorem of Ceva, Menelaus and Van Aubel.

Theorem 4 Let ABC be a triangle with three cevians AA1, BB1 and
CC1 intersecting at a point M (Figure 8).
Furthermore suppose
|A1B|
|A1C|
γ |B1C|
= ,
β |B1A|
Figure 8:
= α
γ
and |C1A|
|C1B|
= β
α .
If X and Y are points on the sides AB and AC then the point M belongs to
the line segment XY if and only if
β( |XB| C|
) + γ(|Y ) = α.
|XA| |Y A|
Proof By van Aubel’s theorem:
|AM|
|A1M|
|C1A| |B1A|
= +
|C1B| |B1C|
= β γ
+
α α
= β + γ
α .
Now suppose M belongs to the line segment XY. Then by the previous lemma
β( |XB| C|
) + γ(|Y )
|XA| |Y A|
= (β + γ)|A1M|
|MA|
= (β + γ)( α
) = α,
β + γ
as required.
For converse, suppose XY and AA1 intersect in point M ′ . We will show that
M ′ coincides M.
By the lemma,
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